Question

How do you solve using the square method $9{x^2} - 12x + 5 = 0$?

Answer 1

Here we are given a quadratic equation. We will be applying completing the square method to find the variable x. Firstly, we take the constant term to R.H.S. of the equation. Then add 4 to both the sides of the quadratic equation. Then make rearrangement in the left hand side to make it complete square. Then we take the square root on both sides and simplify the equation. Then we try to solve it to get the value of the unknown variable x.

Complete step by step answer:
Given the equation $9{x^2} - 12x + 5 = 0$
We have to solve the given quadratic equation by completing the square method.
Let us consider the quadratic equation $9{x^2} - 12x + 5 = 0$ ……(1)
To simplify the equation given above we transfer 5 to the R.H.S. of the equation.
Remember that when transferring any number or variable to the other side, the signs of the same will be changed to the opposite sign.
Take 5 to the R.H.S. in the equation (1), we get,
$\Rightarrow 9{x^2} - 12x = - 5$ …… (2)
Now we add the number 4 to both the sides of the equation (2), we get,
$\Rightarrow 9{x^2} - 12x + 4 = - 5 + 4$
$\Rightarrow 9{x^2} - 12x + 4 = - 1$
Now we write the term on the L.H.S. in the form ${a^2} - 2ab + {b^2}$.
Hence we get,
$\Rightarrow {(3x)^2} - 2 \times 3x \times 2 + {2^2} = - 1$
We have the formula, ${a^2} - 2ab + {b^2} = {(a - b)^2}$
Here $a = 3x$ and $b = 2$
Hence we have,
$\Rightarrow {(3x - 2)^2} = - 1$
Taking square root on both sides we get,
$\Rightarrow \sqrt {{{(3x - 2)}^2}} = \sqrt { - 1}$
We observe that on the L.H.S. we obtain a negative square root.
We know that $\sqrt { - 1} = i$, which is the imaginary unit.
Hence, we get,
$\Rightarrow \sqrt {{{(3x - 2)}^2}} = \pm i$
$\Rightarrow 3x - 2 = \pm i$
Note that we have two solutions.
$\Rightarrow 3x - 2 = i$ and $3x - 2 = - i$
When $3x - 2 = i$, we have,
$\Rightarrow 3x = 2 + i$
Dividing by 3 we get,
$\Rightarrow \dfrac{{3x}}{3} = \dfrac{{2 + i}}{3}$
$\Rightarrow x = \dfrac{{2 + i}}{3}$
When $3x - 2 = - i$, we have,
$\Rightarrow 3x = 2 - i$
Dividing by 3 we get,
$\Rightarrow \dfrac{{3x}}{3} = \dfrac{{2 - i}}{3}$
$\Rightarrow x = \dfrac{{2 - i}}{3}$

Hence the solution of the equation $9{x^2} - 12x + 5 = 0$ by completing square method is given by$x = \dfrac{{2 + i}}{3}$ and $x = \dfrac{{2 - i}}{3}$ .

Note: Here many students solve the equation by using the quadratic formula given by
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
Since it is mentioned, they must keep in mind that they have to solve this kind of question by completing the square method.
Method for completing the square is explained below.
To solve the quadratic equation of the form $a{x^2} + bx + c = 0$ by completing the square:
(1) If a, the leading coefficient is not equal to 1, then divide the whole equation by a.
(2) Transform the quadratic equation so that the constant term c is alone on the right hand side.
(3) Add the square of half the coefficient of x term, ${\left( {\dfrac{b}{{2a}}} \right)^2}$to both sides of the equation.
(4) Factor the left hand side as the square of the binomial.
(5) At last solve for the variable x.